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A tank is full of water. Find the work required to pump the water out of the spout. In Exercises 25 and 26 use the fact that water weighs $ 62.5 lb/ft^3 $.

$1.06 \times 10^{6} \mathrm{J}$

Applications of Integration

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okay to cattle in the work. We need to figure out several things. The volume, the MASS force and here were given the business the difference between the starting starting configuration and the ending impregnation of all these quantities. So just calculated. And no, we're gonna list everything here and we're gonna figure out our work so that the difference of bottom is eight sons. Why time stowed away And, uh, the difference of mass? It's 1000 density times over. This is a volatile. It's equals to 1000 white the other way and the difference of force. So everything is in the consulate shoulder, right. We have a volume and we have the mess because Master Thisted volume and then we have the force, which was two mg. Angie's grab here and the force equals two none. 18 tons. L word, the change of mass. So it's 78,400. Why b y. And that's everything we need to calculate the force and the forest is gonna be the change of the work. The change in war is gonna be the chin of force times five months. Why? What happens over here? Does the water the words really bond ads. Disposition is this height at five Minutes boy, and that's and that's 78,400. Why b y. Why dad away times Who's why comes five minus y the other way? Okay, it's over. Work equals to 0 to 3. Some of the 8400 five. Why tons by minus y times, boy. Five minus y times. Why be Roy? And that's gonna be 78. 470,400 times that or two white squared minus Why people were three 0 to 3 and that's roughly 1.6 comes Tintin. Six. Jule doesn't work.

University of Illinois at Urbana-Champaign

Applications of Integration